# 21 is a discrete analog of this remark for nonlinear

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Unformatted text preview: ty. 4.6 Lyapunov Functions A continuous function is a strong Lyapunov function for an equilibrium of a flow on if there exists an open neighborhood of such that for and for all and . is a weak Lyapunov function if the same holds except and . for all is positive definite on a neighborhood of if and .A strong Lyapunov function is positive definite in some neighborhood of an equilibrium point . Theorem 4.7 (Lyapunov Functions) Let be an equilibrium point of a flow . If is a weak Lyapunov function in some neighborhood of , then is stable. If is a strong Lyapunov function, then is asymptotically stable. Proof. First prove that Let and is stable if is a weak Lyapunov function. Choose so small that Then . Choose so small that . Then if , for all This means is stable. Next prove that is asymptotically stable if is a strong Lyapunov function. Note that is still stable as argued above. Consider a trajectory whose initial point is any . We will show that is the only limit point of the flow. Together with the fact that is stable, this shows that . 12 Chapter 4A Finally prove that is the only limit point of the flow. Consider any sequence of times . Then has a convergent subsequence. For simplicity, denote the subsequence of times too. Suppose . Because is a decreasing function . However, because is not an equilibrium point, continuity of , for for any . Fix large enough, we must also have , By . But then for , which contradicts the chain of inequalities displayed above. □ Example. Show that the origin is an asymptotically stable state for the system , . For a candidate Lyapunov function, try a very simple positive definite function with minimum at the origin. Let . Then by the chain rule for . is a strong Lyapunov function. The origin is an asymptotically stable equilibrium. ■ Recall Hamiltonian systems. Let and and consider . often represents the energy of a system. It is constant along trajectories: . If is an equilibrium, then Is zero at the equilibrium and constant along trajectories. If is positive definite in some neighborhood of the equilibrium, then it is a weak Lyapunov function. Example. The simple pendulum. The Hamiltonian is , , leading to 13 Chapter 4A . sketch support, rod, bob, angle is an equilibrium point. . Consider . is positive definite in a neighborhood of the origin, so Conclude that is a stable equilibrium point. is a weak Lyapunov function. sketch phase portrait near origin. Nearly circular orbits surround the origin. The linearized system is a center. ■ Example. The damped pendulum. The pendulum experiences friction proportional to velocity. , . We now expect the system to lose energy. The system is no longer Hamiltonian. an equilibrium point. Is still a Lyapunov function? is still . Since when , is still a weak Lyap nov f n tion. Lyap n...
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## This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .

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